Gen.Physics-1-2nd-Quarter-Final
Definition: Is the constant speed a freely falling object reaches when the resistance of the medium (like air or water) prevents further acceleration.
Key Points:
At terminal velocity, the forces are balanced:
The force of gravity pulling the object down is equal to the force of air resistance acting upwards.
The object ceases to accelerate and continues falling at a constant velocity.
Formula:
( F_{grav} = mg ) Where:
( m ): mass of the object (kg)
( g ): acceleration due to gravity (approx. 9.8 m/s² on Earth)
Determine the velocity of a rock dropped from a bridge with an initial velocity of 15 m/s after 3.0 seconds.
Initial velocity (( v_0 )): 15 m/s [down] ((=-15 m/s))
Acceleration (( a )): ( g = 9.8 , \text{m/s}^2 ) [down] ((=-9.8 , \text{m/s}^2))
Time (( , t = 3.0 , s ))
Calculate final velocity:
( v = v_0 + a imes t )
( v = -15 , m/s + (-9.8 , m/s^2)(3.0 , s) )
( v = -15 - 29.4 , m/s )
( v = -44.4 , m/s ) or 44 m/s [down]
Determine the velocity of a baseball thrown upwards with an initial velocity of 15 m/s after 2.0 seconds.
Initial velocity (( v_0 )): 15 m/s [up] ((=15 m/s))
Acceleration (( a )): (=-9.8 , m/s^2) [down]
Time (( , t = 2.0 , s ))
Calculate final velocity:
( v = v_0 + a imes t )
( v = 15 , m/s + (-9.8 , m/s^2)(2.0 , s) )
( v = 15 - 19.6 , m/s )
( v = -4.6 , m/s )
A simple pendulum consists of an object (the bob) attached to a string that swings back and forth.
Amplitude: The maximum displacement from the the pendulum's rest position when swinging.
Period: The time needed for the pendulum to make one complete cycle of motion.
Escape Velocity is the minimum speed required for an object to break free from a planet's gravitational pull without additional propulsion.
At this speed, an object possesses sufficient energy to escape the gravitational attraction and travel away indefinitely.
Example: The speed required for a spacecraft to leave Earth.
Describes that planetary orbits are elliptical, not circular, with the Sun located at one focus of the ellipse.
States that planets move faster when closer to the Sun (perihelion) and slower at greater distances (aphelion).
The area swept out over equal times remains constant.
Indicates that the time taken for a planet to complete an orbit increases with distance from the Sun.
For instance, Neptune has a far longer orbital period than Earth due to its greater distance from the Sun.
The force that attracts objects with mass towards each other.
On Earth, it keeps us grounded and causes dropped objects to fall.
Calculate the force of gravity for a car with a mass of 1,250 kg:
( F = mg = (1,250 kg)(9.8 N/kg) = 12,250 N )
Mass: Measured in kilograms (kg); the amount of matter in an object.
Weight: Measured in Newtons (N); the force of gravity acting on an object.
The gravitational force depends on the masses of the objects involved and the distance between them. Larger masses create stronger gravitational pulls.
( F = rac{m_1 m_2}{r^2} )
F: Gravitational force (N)
m1, m2: Masses of the objects (kg)
r: Distance between the objects (m)
G: Gravitational constant ( G = 6.67 imes 10^{-11} , N , m^2/kg^2 )
Determine the gravitational force between two objects of equal mass (0.50 kg) that are 10 cm apart.
Masses (( m_1 = m_2 = 0.50 kg ))
Distance (10 cm = 0.10 m)
Gravitational Constant (( G = 6.67 imes 10^{-11} , N , m^2/kg^2 ))
( F = G rac{m_1 m_2}{r^2} )
( F = 6.67 imes 10^{-11} imes rac{(0.50)(0.50)}{(0.10)^2} = 1.7 imes 10^{-9} N )
Calculate the gravitational force between two objects with masses of 10 kg and 20 kg that are 2 meters apart.
Determine the gravitational force between Earth and Moon given:
Mass of Earth = ( 5.97 imes 10^{24} kg )
Mass of Moon = ( 7.35 imes 10^{22} kg )
Average distance between them = ( 3.84 imes 10^8 m )
Hooke's Law states that the force needed to stretch or compress a spring is proportional to the distance it is stretched or compressed within its elastic limit.
( F = kx )
F: Force applied to the spring (N)
k: Spring constant (measure of stiffness)
x: Displacement from natural position (m)
Given: Spring constant ( k = 200 , N/m ), displacement ( x = 0.15m )
Calculation:
( F = kx = (200 , N/m)(0.15 , m) = 30N )
Given: Spring constant ( k = 48 , N/m ), mass = 0.25 kg.
Calculation:
Weight (Force) ( F = (0.25 , kg)(9.8 , N/kg) = 2.4 , N )
Displacement ( x = rac{F}{k} = rac{2.4 , N}{48 , N/m} = 0.050 , m , (5.0 , cm) )
Friction is the force opposing the relative motion or tendency between two surfaces in contact.
Static Friction:
Resists the start of motion (e.g., a book on a table).
Kinetic Friction:
Opposes motion between sliding surfaces (e.g., sledding down a hill).
Rolling Friction:
Resists rolling motion (e.g., wheels on a road).
Fluid Friction:
Resistance an object experiences when moving through a fluid (e.g., swimming).
The overall force acting on an object when all forces are combined.
Determines if an object remains stationary, moves constantly, or accelerates.
Determine the resultant force on a block of wood with forces of 10 N [East] and 20 N [East].
Answer: 30 N [East].
Determine the net force on a rope if one pulls with 20 N [East] and another pulls with 5 N [West].
Answer: 15 N [East].
Definition: Is the constant speed a freely falling object reaches when the resistance of the medium (like air or water) prevents further acceleration.
Key Points:
At terminal velocity, the forces are balanced:
The force of gravity pulling the object down is equal to the force of air resistance acting upwards.
The object ceases to accelerate and continues falling at a constant velocity.
Formula:
( F_{grav} = mg ) Where:
( m ): mass of the object (kg)
( g ): acceleration due to gravity (approx. 9.8 m/s² on Earth)
Determine the velocity of a rock dropped from a bridge with an initial velocity of 15 m/s after 3.0 seconds.
Initial velocity (( v_0 )): 15 m/s [down] ((=-15 m/s))
Acceleration (( a )): ( g = 9.8 , \text{m/s}^2 ) [down] ((=-9.8 , \text{m/s}^2))
Time (( , t = 3.0 , s ))
Calculate final velocity:
( v = v_0 + a imes t )
( v = -15 , m/s + (-9.8 , m/s^2)(3.0 , s) )
( v = -15 - 29.4 , m/s )
( v = -44.4 , m/s ) or 44 m/s [down]
Determine the velocity of a baseball thrown upwards with an initial velocity of 15 m/s after 2.0 seconds.
Initial velocity (( v_0 )): 15 m/s [up] ((=15 m/s))
Acceleration (( a )): (=-9.8 , m/s^2) [down]
Time (( , t = 2.0 , s ))
Calculate final velocity:
( v = v_0 + a imes t )
( v = 15 , m/s + (-9.8 , m/s^2)(2.0 , s) )
( v = 15 - 19.6 , m/s )
( v = -4.6 , m/s )
A simple pendulum consists of an object (the bob) attached to a string that swings back and forth.
Amplitude: The maximum displacement from the the pendulum's rest position when swinging.
Period: The time needed for the pendulum to make one complete cycle of motion.
Escape Velocity is the minimum speed required for an object to break free from a planet's gravitational pull without additional propulsion.
At this speed, an object possesses sufficient energy to escape the gravitational attraction and travel away indefinitely.
Example: The speed required for a spacecraft to leave Earth.
Describes that planetary orbits are elliptical, not circular, with the Sun located at one focus of the ellipse.
States that planets move faster when closer to the Sun (perihelion) and slower at greater distances (aphelion).
The area swept out over equal times remains constant.
Indicates that the time taken for a planet to complete an orbit increases with distance from the Sun.
For instance, Neptune has a far longer orbital period than Earth due to its greater distance from the Sun.
The force that attracts objects with mass towards each other.
On Earth, it keeps us grounded and causes dropped objects to fall.
Calculate the force of gravity for a car with a mass of 1,250 kg:
( F = mg = (1,250 kg)(9.8 N/kg) = 12,250 N )
Mass: Measured in kilograms (kg); the amount of matter in an object.
Weight: Measured in Newtons (N); the force of gravity acting on an object.
The gravitational force depends on the masses of the objects involved and the distance between them. Larger masses create stronger gravitational pulls.
( F = rac{m_1 m_2}{r^2} )
F: Gravitational force (N)
m1, m2: Masses of the objects (kg)
r: Distance between the objects (m)
G: Gravitational constant ( G = 6.67 imes 10^{-11} , N , m^2/kg^2 )
Determine the gravitational force between two objects of equal mass (0.50 kg) that are 10 cm apart.
Masses (( m_1 = m_2 = 0.50 kg ))
Distance (10 cm = 0.10 m)
Gravitational Constant (( G = 6.67 imes 10^{-11} , N , m^2/kg^2 ))
( F = G rac{m_1 m_2}{r^2} )
( F = 6.67 imes 10^{-11} imes rac{(0.50)(0.50)}{(0.10)^2} = 1.7 imes 10^{-9} N )
Calculate the gravitational force between two objects with masses of 10 kg and 20 kg that are 2 meters apart.
Determine the gravitational force between Earth and Moon given:
Mass of Earth = ( 5.97 imes 10^{24} kg )
Mass of Moon = ( 7.35 imes 10^{22} kg )
Average distance between them = ( 3.84 imes 10^8 m )
Hooke's Law states that the force needed to stretch or compress a spring is proportional to the distance it is stretched or compressed within its elastic limit.
( F = kx )
F: Force applied to the spring (N)
k: Spring constant (measure of stiffness)
x: Displacement from natural position (m)
Given: Spring constant ( k = 200 , N/m ), displacement ( x = 0.15m )
Calculation:
( F = kx = (200 , N/m)(0.15 , m) = 30N )
Given: Spring constant ( k = 48 , N/m ), mass = 0.25 kg.
Calculation:
Weight (Force) ( F = (0.25 , kg)(9.8 , N/kg) = 2.4 , N )
Displacement ( x = rac{F}{k} = rac{2.4 , N}{48 , N/m} = 0.050 , m , (5.0 , cm) )
Friction is the force opposing the relative motion or tendency between two surfaces in contact.
Static Friction:
Resists the start of motion (e.g., a book on a table).
Kinetic Friction:
Opposes motion between sliding surfaces (e.g., sledding down a hill).
Rolling Friction:
Resists rolling motion (e.g., wheels on a road).
Fluid Friction:
Resistance an object experiences when moving through a fluid (e.g., swimming).
The overall force acting on an object when all forces are combined.
Determines if an object remains stationary, moves constantly, or accelerates.
Determine the resultant force on a block of wood with forces of 10 N [East] and 20 N [East].
Answer: 30 N [East].
Determine the net force on a rope if one pulls with 20 N [East] and another pulls with 5 N [West].
Answer: 15 N [East].